Question: $h(x)=-(x+11)^2+1$ 1) What are the zeros of the function? Write the smaller $x$ first, and the larger $x$ second. $\text{smaller }x=$
Explanation: $\begin{aligned} -(x+11)^2+1&=0 \\\\ (x+11)^2&=1 \\\\ \sqrt{(x+11)^2}&=\sqrt{1} \\\\ x+11&=\pm 1 \\\\ x&=\pm1-11 \\\\ x={-12}&\text{ or }x={-10} \end{aligned}$ $h(x)$ is given in vertex form: $h(x)=-(x-({-11}))^2+{1}$ So the vertex of the parabola is at $({-11},{1})$. In conclusion, $\begin{aligned} \text{smaller }x&=-12 \\\\ \text{larger }x&=-10 \end{aligned}$ The vertex of the parabola is at $(-11,1)$